1. Optimization Models
Module 9

2. OPTIMIZATION MODELS
EXTERNAL
INPUTS
OUTPUT
MODEL
DECISION
INPUTS
Optimization models answer the question, “What decision values give the best outputs?”

3. SCARCITY
DECISION
INPUTS
CONSTRAINTS
Optimization models are necessary because resources are always limited.

4. LINEAR PROGRAMMING
MODEL
(objective
function)
10x + 18y
DECISION
INPUTS
CONSTRAINTS
OUTPUT
2x + 6y < 150 x
x y < 40
x y < 50 y
Linear programming is a special type of optimization model that sets up constraints as linear equations and solves them simultaneously while maximizing an objective function.

5. EXAMPLE 1: BOB’S MUFFLER SHOP
Bob manufactures 2 types of mufflers, family and sport. Each family muffler nets him $10 and each sport muffler nets him $18. These numbers are known as the products’ ‘contribution margin’.
His profit is determined by the function $10f + $18s, where f and s are the total number of each type of muffler sold. This is his objective function.
His decision variables are the number of family and sport mufflers to manufacture each month.
In an unconstrained situation, he would want to sell as many sport mufflers as because of their higher contribution. However…

6. EXAMPLE 1: BACKGROUND
In order to create a muffler, Bob must use brackets, labor hours, and alloy. The resources needed for each type are:
Family
Sports
Brackets
2.00
6.00
Labor hours
1.00
0.80
Max Alloy 1
1.5
In any given month, he is limited in all three resources to these amounts:
Quantity
Brackets
150
Labor hours 40
Max Alloy 50
These are the problem constraints.

7. EXAMPLE 1: MODELING CONSTRAINTS AS LINEAR EQUATIONS
These constraints can be modeled as a system of linear equations:
2f s ≤ (brackets)
f s ≤ (labor)
50f s ≤ (alloy)
Bob also has an outstanding contract that forces him to manufacture 5 sports style mufflers each month and knows, from historical performance, that demand for family mufflers in a given month will not exceed 35.
These constraints can be modeled as:
f ≤ (maximum demand)
s ≥ (contractual obligation)
The objective function (profit) is:
10f + 18s

8. EXAMPLE 1: GRAPHING A CONSTRAINT
This situation can be represented graphically. Consider the constraint imposed by brackets: 2f + 6s ≤ 150.
If zero family mufflers are manufactured, the equation changes to (0)f + 6s ≤ 150, which reduces to s ≤ 25. Thus a maximum of 25 sports mufflers can be made if no family mufflers are made.
Similarly, 75 family mufflers can be made if there are no sports mufflers manufactured.
Connecting these points gives a graphical representation of this constraint.

9. EXAMPLE 1: RANGE OF FEASIBILITY
Points on the line are scenarios in which all available brackets are used.
The area under the line is called the Range of Feasibility because it covers all product mixes that the constraint allows.
Any point above the line is infeasible because Bob does not have enough brackets to create that combination of mufflers.
Infeasible
Point:
(30 sport,
40 family)
Range of
Feasibility

10. EXAMPLE 1: GRAPHING ALL CONSTRAINTS
The rest of the constraints can then be added in a similar fashion.
2f + 6s ≤ (brackets)
f + .8s ≤ (labor)
f + 1.5s ≤ (alloy)
f ≤ (maximum demand)
s ≥ (contractual obligation)

11. EXAMPLE 1: THE COMPOSITE RANGE OF FEASIBILITY
The Range of Feasibility for the model is the area enclosed by all lines.
2f + 6s ≤ (brackets)
f + .8s ≤ (labor)
f + 1.5s ≤ (alloy)
f ≤ (maximum demand)
s ≥ (contractual obligation)
Range of
Feasibility

12. EXAMPLE 1: POTENTIAL PROFIT LINES
The profit function will be a line with the slope of -5/9, the ratio of the contribution margin of family mufflers to sports mufflers.
Each line corresponds to a particular level of total profit. The further out the line is, the greater the amount of profit.
Total Profit:
$ $ $ $ $ $550

13. EXAMPLE 1: FEASIBLE PROFIT LINES
By combining the profit lines with the range of feasibility, you can identify the highest profit line that touches a feasible point.

14. EXAMPLE 1: OPTIMAL SOLUTION VIA GRAPHICAL METHOD
The optimal point is a mix of 25 family mufflers and sport mufflers. This combination will net Bob $550.
Optimal
point

15. EXAMPLE 1: SPREADSHEET MODEL
The
situation
can be
modeled
in Excel
and an
optimal
solution
found
with
Solver.

16. EXAMPLE 1: SPREADSHEET MODEL
Contribution
Margins
Available resources
Resources
needed
for each
type of
muffler
Profit
Max
demand
Contract
Decisions:
How many
of each
muffler to
make
Modeled constraints

17. EXAMPLE 1: SOLVER INPUTS
Select the profit cell and choose solver from the tools menu.

18. EXAMPLE 1: SOLVER INPUTS
Objective
function (profit)
Click
Decision
variables
Constraints

19. EXAMPLE 1: SOLVER RESULTS
Solver finds a solution. Select the answer and sensitivity reports for further analysis.

20. EXAMPLE 1: OPTIMAL SOLUTION
Solver found the same optimal solution
as the graphical method.

21. EXAMPLE 1: ANSWER REPORT
‘Binding’ means
all available
resources are
used
Brackets: 2f + 6s ≤ 150
Labor: f + .8s ≤ 40
The total
resources
used in the
optimal
solution
Slack is
the amount
of leftover
resources
All available brackets are used in the optimal solution, making it a binding constraint. There are 1.67 labor hours left over, or slack.

22. EXAMPLE 1: SENSITIVITY ANALYSIS
The shadow price is the amount of profit that an additional unit of available resources would yield.
A non-binding constraint will have a shadow price of 0 because there is already an excess of the resource.
An additional available bracket (a total of 151) will yield $1 of profit. An additional labor hour will yield no profit because there is already an excess.
The allowable increase is the amount of resources that would have to be available to change the optimal solution.
If 50 additional brackets are available, the optimal product mix will change.
The allowable decrease is the amount of fewer resources that will change the optimal mix.

23. EXAMPLE 1: SENSITIVITY ANALYSIS
If an additional unit of resource can be obtained for less than the shadow price, then the difference will be profit.
The allowable increase and decrease tells you how sensitive the decision inputs of the optimal solution (final value column) are to the right hand sides of each constraint.

24. EXAMPLE 1: RANGE OF OPTIMALITY
Profit: 10 f + 18 s
The contribution margin for each type of muffler may be uncertain.
The range of optimality is the range of contribution margins (coefficients in the objective function) in which a particular solution is the optimal solution.

25. EXAMPLE 1: RANGE OF OPTIMALITY
Profit: 10 f + 18 s
8 ≤ contribution margin ≤ 14
15 ≤ contribution margin ≤ 30
Sports mufflers range:
$18 + $12 = $30
$18 - $3 = $15
The range of optimality tells you how sensitive the decision inputs for the optimal solution are to variation in the coefficients of the objective function.